Solution of the initial value problem in partial differential equation

Question

The solution to the Initial value Problem $(x-y)\frac{\partial u}{\partial x} +(y-x-u)\frac{\partial u}{\partial y} =u$ and $u(x,0)=1$ satisfies

1. $u^2(x-y+u)+(y-x-u)=0$
2. $u^2(x+y+u)+(y-x-u)=0$
3. $u^2(x-y+u)-(x+y+u)=0$
4. $u^2(y-x+u)+(x+y-u)=0$

after solving this linear partial differential equation one solution is $x+y+u=a$. Can someone help me with this problem. Thanx in Advance

Answer 1

First checking :

1. $u^2(x-y+u)+(y-x-u)=0 \quad\to\quad 1^2(x-0+1)+(0-x-1)=0$
2. $u^2(x+y+u)+(y-x-u)=0 \quad\to\quad 1^2(x+0+1)+(0-x-1)=0$
3. $u^2(x-y+u)-(x+y+u)=0 \quad\to\quad 1^2(x-0+1)-(x+0+1)=0$
4. $u^2(y-x+u)+(x+y-u)=0 \quad\to\quad 1^2(0-x+1)+(x+0-1)=0$

In the four cases the condition $u(x,0)=1$ is satisfied. This don't allow us to eliminate a case among the four. So, we now have to check the agreement to the PDE for the four cases, one after the other.

They are two methods :

FIRST METHOD : Example with the first case : $u^2(x-y+u)+(y-x-u)=0$

The implicit differentiation leads to : $$\frac{\partial u}{\partial x} = \frac{-u^2+1}{2u(x-y+u)+u^2-1}$$ $$\frac{\partial u}{\partial y} = \frac{u^2-1}{2u(x-y+u)+u^2-1}$$ The condition $u(x,0)=1$ leads to $\frac{\partial u}{\partial x}(x,0) =\frac{\partial u}{\partial y}(x,0) =0$ . Puting it into the PDE shows the contradiction $0=1$. Conclusion : The case $1$ is rejected.

I let you continue up to finding the case which is not rejected.

SECOND MEHOD :

Solving the PDE in order to express the general solution. Then comparing to the four proposed solutions and determining which one agrees.

INT : Thanks to the method of characteristics, the general solution expressed on the form of an implicite equation is :

$$F\left( (x+y+u)\:,\:\frac{u^2}{x-y+u}\right)=0$$ where $F$ is any differentiable function of two variables.

This allows to determine which case agrees among the four proposed.